I recall that in my first year of junior high, I learned a method for drawing an ellipse from a friend who was an art student: select a rectangle, take the midpoints of a pair of adjacent sides, connect them, and cut off the resulting triangle. In the remaining pentagon, continue taking the midpoints of adjacent sides, connect them, and cut them off, obtaining a figure as shown below. Then, draw an arc that is as tangent as possible to AG, GH, HI, and IJ in the figure; this arc will roughly be a quarter of an ellipse.
I used to think that by infinitely halving the sides in this way, one could obtain a precise ellipse. However, I soon refuted this conclusion. The reason is simple: consider the figure below (a square with side length 1). If my conclusion were correct, this figure should converge to a quarter circle, and its arc length should converge to \frac{\pi}{2}. However, using the Pythagorean theorem to calculate, one finds that the sum of the side lengths of the resulting polygon (excluding the two longest legs) quickly becomes less than \frac{\pi}{2}.
So, what shape does the limit of this halving process yield? I originally thought it would result in an interesting curve, but calculations from a few days ago showed me that it is not an exciting curve at all—the final result is—Triangle ABC! Yes, this cutting method actually manages to cut away half of the square. (Fortunately, my friend only performed a finite number of cuts.) Let us try to prove this conclusion below.
First, let’s look at the distribution pattern of these points. Let point D be the origin, point A be (1,0), and point C be (0,1). Then we can list the following table:
This is a numerical array similar to Pascal’s triangle, where each coordinate is the average of the two adjacent coordinates above it; the ends are of the form (\frac{2^n-1}{2^n}, 0). To see the pattern more clearly, let’s look only at the x-coordinates:
It is not difficult to see that the denominators of the numbers in the (n+1)-th row are all 2^n, and the difference between two adjacent numerators is exactly the n-th row of Pascal’s triangle. That is, the i-th number of the (n+1)-th row is: \frac{2^n - C_n^0 - C_n^1 - \dots - C_n^{i-1}}{2^n}
To prove that the final curve tends toward a straight line, we only need to prove that the midpoint coordinate tends toward (1/2, 1/2). To this end, we consider the numbers in the (2n+1)-th row, where the midpoint coordinate is (k, k), with: k = \frac{2^{2n} - C_{2n}^0 - C_{2n}^1 - \dots - C_{2n}^n}{2^{2n}}
According to C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + \dots + C_{2n}^{2n} = 2^{2n}, k can be rewritten as: k = \frac{2^{2n} - \frac{2^{2n} + C_{2n}^n}{2}}{2^{2n}} = 1/2 - \frac{C_{2n}^n}{2^{2n+1}}
Now the question is whether \frac{C_{2n}^n}{2^{2n+1}} tends to 0 as n tends to infinity. BoJone used a scientific calculator to calculate for n=50 and found the result to be 0.036\dots This did not show signs of convergence, but the calculator’s limit was n=50. Thus, there was no choice but to find another way. Since C_{2n}^n = \frac{(2n)!}{(n!)^2}, by Stirling’s formula: \lim_{n\to \infty} \frac{\sqrt{2\pi n} (\frac{n}{e})^n}{n!} = 1
Thus (n!)^2 \approx 2\pi n (\frac{n}{e})^{2n} and (2n)! \approx 2\sqrt{\pi n} (\frac{2n}{e})^{2n}, which means: C_{2n}^n \approx 2^{2n} \cdot \frac{1}{\sqrt{\pi n}}
Finally, we obtain: \lim_{n\to \infty} \frac{C_{2n}^n}{2^{2n+1}} = \lim_{n\to \infty} \frac{1}{2 \sqrt{\pi n}} = 0
Strictly speaking, using the equals sign in the above derivation is not entirely correct. However, since we are only seeking the limit and limits are transitive, writing it this way poses no major problem. Thus, we have proven that the midpoint of the figure after cutting tends toward (1/2, 1/2), meaning the final result is a straight line!
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